# A sample of a gaseous compound

A sample of a gaseous compound of nitrogen and oxygen weighs5.25 g and occupies a volume of 1.00 L at a pressure of 1.26 atmand temperature of -4.0 °C. Which of the following molecularformulas could be that of the compound?

Answer is N_{2}O_{4}

_{Please show work, thank you!}

Answer:

Ideal Gas Equation: PV = nRT :

We have Pressure P = 1.26 atm;Volume V = 1LR = 0.08205 L atm / mol·KT = -4.0^{o}C = -4.0 + 273 = 269K

(1.26 atm) (1.00 L) = (n) (0.08205 L atm / mol·K ) (269 K)

n = 0.0570 mol

Now, we have no. moles and also given the mass of the gas m =5.25 gTherefore, we calculate the molecular weight:no. of moles = mass/mol.wtmol.wt = mass/no.moles = 5.25 g / 0.057 mol = 92.1 g/mol

Now we have to find the empirical formula, but there is nofurther information about the how much % of each elementcontributes. Therefore, the information is not complete in thisproblem.We need mass percent of each element in the gas to determine theempirical formula.From the empirical formula and molecular weight, you can determinethe molecular formula.

Suppose, the contains 30.6% nitrogen and 69.4% oxygen bymass,

Let’s assume 100 g of the gas is present.

N: 30.6g/14.007g/mol = 2.1846 molO: 69.4g / 16.00g/mol = 4.3375 mol

N: 2.1846/2.1846 = 1O: 4.3375/2.1846 = 1.985 = 2

The Empirical formula = NO_{2}

Empirical formula wt = 14 + 2(16) = 46

Therefore, molecular wt/empirical formula wt= 92.1 / 46 ~2Therefore the molecular formula is : (NO_{2})_{2} =N_{2}O_{4}